Problem: Simplify; express your answer in exponential form. Assume $k\neq 0, a\neq 0$. $\dfrac{{(k^{-4}a^{5})^{-1}}}{{(k^{-1}a^{-5})^{2}}}$
Solution: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(k^{-4}a^{5})^{-1} = (k^{-4})^{-1}(a^{5})^{-1}}$ On the left, we have ${k^{-4}}$ to the exponent ${-1}$ . Now ${-4 \times -1 = 4}$ , so ${(k^{-4})^{-1} = k^{4}}$ Apply the ideas above to simplify the equation. $\dfrac{{(k^{-4}a^{5})^{-1}}}{{(k^{-1}a^{-5})^{2}}} = \dfrac{{k^{4}a^{-5}}}{{k^{-2}a^{-10}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{4}a^{-5}}}{{k^{-2}a^{-10}}} = \dfrac{{k^{4}}}{{k^{-2}}} \cdot \dfrac{{a^{-5}}}{{a^{-10}}} = k^{{4} - {(-2)}} \cdot a^{{-5} - {(-10)}} = k^{6}a^{5}$